A) 28.6m
B) 8.6m
C) 6m
D) 18.6m
Correct Answer: A
Solution :
The rotational and translational kinetic energy of the ball at the bottom will be changed to gravitational potential energy, when the sphere stops. We therefore write \[{{\left( \frac{M{{v}^{2}}}{2}+\frac{I{{\omega }^{2}}}{2} \right)}_{start}}={{(Mgh)}_{end}}\] For a solid sphere \[I=\frac{2M{{R}^{2}}}{5}\] Also,\[\omega =\frac{v}{r},\]then above equation becomes \[\frac{1}{2}M{{v}^{2}}+\frac{1}{2}\left( \frac{2}{5}M{{r}^{2}} \right){{\left( \frac{v}{r} \right)}^{2}}=Mgh\] \[\frac{1}{2}{{v}^{2}}+\frac{1}{5}{{v}^{2}}=(g+8)h\] Using v = 20 m/s gives h = 28.6 m
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